Question

Find the magnitude of the projection of the vector\[2i + 3j + k\], on a vector which is perpendicular to the plane consisting of vectors \[i + j + k\]and\[i + 2j + 3k\].
\[
  {\text{A}}{\text{.}}\dfrac{{\sqrt {\text{3}} }}{{\sqrt {\text{2}} }} \\
  {\text{B}}{\text{.}}\dfrac{{\sqrt {\text{2}} }}{{\sqrt {\text{3}} }} \\
  {\text{C}}{\text{.}}\dfrac{{\text{4}}}{{\sqrt {\text{3}} }} \\
  {\text{D}}{\text{.}}\dfrac{{{\text{2}}\sqrt {\text{2}} }}{{\sqrt {\text{3}} }} \\
 \]

Answer 1

Hint: A quantity having magnitude as well as direction is known as a vector, necessary for finding the position of one point in space with respect to others. Vectors are represented by a line symbol \[ \to \] on the top of the variable that determines the magnitude and direction of the quantity.
The vector projection is the vector produced where one vector is resolved into two components, one being parallel to the other vector and other being parallel to it. Vector projection of one vector on the other can be found by \[\dfrac{{\vec a \cdot \overrightarrow b }}{{\left| b \right|}}\] where, \[\overrightarrow b \] is a vector on which \[\vec a\] vector is projected.
To check whether the given two vectors are perpendicular, we need to find their scalar, which should be equal to zero, i.e. \[\vec a \cdot \vec b = 0\].
To find the projection of a vector, find the vector on which this vector is projected, which is also a perpendicular plane.

Complete step by step answer:
Let the vector which is perpendicular to the plane be\[\vec z\],
Vectors \[\vec x = \hat i + \hat j + \hat k\] and \[\vec y = \hat i + 2\hat j + 3\hat k\] lie on the plane, now vector \[\vec z\] which is perpendicular to these two vectors will be \[\vec z = \vec x \times \vec y\] their cross product is:
\[
  \vec x \times \vec y = \left| {\begin{array}{*{20}{c}}
  {\hat i}&{\hat j}&{\hat k} \\
  1&1&1 \\
  1&2&3
\end{array}} \right| \\
   = \hat i\left( {3 - 2} \right) - \hat j\left( {3 - 1} \right) + \hat k\left( {2 - 1} \right) \\
   = \hat i - 2\hat j + \hat k \\
 \]
Hence, vector \[\vec z = \hat i - 2\hat j + \hat k\]
Now it is given a vector \[\vec w = 2\hat i + 3\hat j + \hat k\]is projected to vector\[\vec z = \hat i - 2\hat j + \hat k\], so we will find their projection which is given as \[\dfrac{{\vec a \cdot \vec z}}{{|z|}}\]
Hence the projection will be
\[
  {\text{Projection}} = \dfrac{{\vec w \cdot \vec z}}{{|z|}} \\
   = \dfrac{{\left( {2\hat i + 3\hat j + \hat k} \right) \cdot \left( {\hat i - 2\hat j + \hat k} \right)}}{{\left| {\hat i - 2\hat j + \hat k} \right|}} \\
   = \left| {\dfrac{{2 - 6 + 1}}{{\sqrt {1 + 4 + 1} }}} \right| \\
   = \left| {\dfrac{{ - 3}}{{\sqrt 6 }}} \right| \\
   = \dfrac{{\sqrt 3 }}{{\sqrt 2 }} \\
   = \sqrt {\dfrac{3}{2}} \\
 \]

Note: When a vector is projected on another vector we find their dot product. When a given vector makes an angle of \[{90^ \circ }\] with the other vector, then the vector is known as an orthogonal vector.